Kothari D.P., Nagrath I.J. Modern Power Systems Analysis

Tranefer

Fcn

Tranefer

Fcn

1

SS1

Moo"rn

po*b,

syrt"r

Rn"trrri,

.

Ksg

Kps

Tps.s

+

1

Conetant

Fig.

G.3

First

order

approximation

for load

frequency

control

Ex.

G.12

The

system

of

Fig.

8.10

is

simulated

using

simulink

as was

Example

G.l l.

Flg.

G.4

Proportional plus

Integral

load

frequency

control

Ex.

G.13

(Example

9.8)

% Program

for

bui]dirrg

of

Zbus

by

addit'ion

of

branch

or l.ink

% Zprimary=[elementno

from

to value

.s+1

done

in

%

yo

%

--l

% Here

care

should

be

taken

that

to

begin

with

an

%

reference

and both

from

and

to nodes

should

not

cl

ear

zpri

mary=

[

1 I

0

0.25

2 2

I 0.1

3 3 1

0.1

4

2

0

0.25

s

2

3 0.11

[el

ements

co1 umns]

=size(zprimary)

t

To

begin

with

zbus

matrix

is

a null

matrix

element

is

added

be new

nodes

to

Ksg

Tsg.s

+

1

Kps

Tps.s

+

1

Transfer

Fcn

Transfer

Fcnl

I

Transfer

Fcn2

lffiw

ffiflI

I currentbusno

indicates

maximum

no.

of buses

added

until

now

currentbusno=0

for

count=1:elements,

[rows

co1 s]

=sjze(zbus)

f rom=zpri

mary

(count,

2)

to=zpri

mary

(count,3)

val

ue=zpri

mary

(count,

4)

% newbus

variable indicates

% newbus

bus

may

or may not

newbus=max

(from,

to)

the

maximum

of

the

two

buses

al ready

be

a

part

of

exi sti ng

zbus

bus

to reference

bus

% ref

variable indicates

the minimum

of the

two

buses

%

& not necessarjly

the reference

bus

% ref bus

must always

exist

in the

existing

zbus

ref=mi

n

(from,

to)

%

Modjfication

of

typel

%

A

new element

is

added from

i f

newbus

>currentbusno

& ref

zbus=[zbus

zeros(rows,

1)

zeros(1,cols)

valuel

cu rrentbusno=newbu

s

conti

nue

end

new

==0

%

Modification

of type?

%

A

new

element is

added

from new

bus

to

old bus

bus

i f newbus

>currentbusno

& ref

-=0

zbus=[zbus

zbus(

:,ref)

zbus

(ref,

:

)

val

ue+zbus

(ref,

ref)]

cu rrentbu

sno=newbu

s

conti

nue

\

other

than

reference

end

% Modi fi

cai ton of type3

% A new

element is

added

between

an

old

bus

and reference

bus

i f newbus

<=currentbusno

& ref==O

zbus=zbus-1/(zbus

(newbus,

newbus)+val

ue)

*zbus

(:,

newbus)

*zbus

(newbus,

:

)

conti nue

end

% Mod'ificatjon

of type4

%

A new

element is

added

betwen

two

old buses

i f

newbus

<=

currentbusno

& ref

-=0

zbus=zbus

-

L/

(value+zbus

(from,

from)+zbus

(to,to)

-

2*zbus(from,to))*((zbus(:,from)-zbus(:,to))*((zbus(from,;|-zbuS(tor:))))

conti

nue

end

Wl

uodgin Power System

Rnatysis

Ex.

G.14

(Example

12,10)

t Program

for

trans'ient stabi'lity of single

machine connected

to lnfinite

%

bus this

program

simulates Example

12.10 usjng

point

by

point

method

cl

ear

t-0

tf=0

tfi

nal

=0.5

tc=O. 125

tsteP=9.

95

l4=?

.5?

/(

180*50)

j=2

de1 ta=21..64*pi

/180

ddel

ta=0

time

(

1)

=9

ang(1)=2I.64

Pm=0.9

Pnaxbf=2.44

Pmaxdf=0.88

Pmaxaf=2.00

whi I

e t<tfi nal

,

'if

(t==tf

),

Pam'i

nus=0.9-Pmaxbf*si

n

(del

ta)

Papl us=O.9-Pmaxdf*s'in

(del

ta)

Paav=

(

Pam'i nus+Pap1

us

)

/2

'

Pa=Paav

end

i f

(t==tc),

Pami

nus=0.9-Pmaxdf*si

n

(del

ta)

Papl us=0. 9-Pmaxaf*s

j

n

(del

ta)

Paav=

(Pamj

nus+P aplus)

/2

Pa=Paav

end

if(t>tf

&

t<tc),

Pa=Pm-Pmaxdf*si n

(del

ta)

end

i f(t>tc),

Pa=Pm-Pmaxaf*si

n

(del

ta)

end

t, Pa

ddel ta=cidel

ta+

(tstep*tstep*pa/M)

del

ta-

(del

ta*180/pj

+ddel

ta)

*pi

/180

de'l tadeg=de1 ta*180/P'i

t=t+tstep

pause

time(i

)=1

ang(i)=deltadeg

j=j+1

end

axis([0

0.6

0 160])

Ex. G.15

Here the

earlier

Example

G.l4

is solved

again using SIMULINK.

Before running

simulation

shown

in Fig. G.5

integrater t has to be

initialized

to

prefault

value

of

4

i.e.

4.

Thit can be

done by double-clicking on

integrater

1 block and

changing

the initial

value

from 0 to 6s

(in

radius). Also double click

the switch

block

and change

the threshold

value from 0 to the fault

clearing

time

(in

sec.).

Ex. G.16

(Ex.

12.11)

v"

Th'ts

program

simul

ates

transi ent stab'i

1 i ty of

mul timachi ne systems

% The data

is from Examp'le

12.11

cl

ear al

I

format long

%Step

1 In'itialisation

with load flow and

mach'ine data

f=50;tstep=9.91.

6=[12

9J,

;

Pgnetterm=[

3.25

2.10]'

;

Qgnetterm=[

0.6986

0.3110]'

;

Xg=[0.067 0.10]';

% Note the

.use

of .'

operator

here

%

This does

a

transpose

wjthout

taking the conjugate

of each element

yg=[po'larTorect(1.03,

8.235)

polarforect(1.02,

7.16)]

.'

% n is

no of

generators

other

than sl ack

bus

n=2i

%Step

2

V0coni=cori

(V0);

Ig0=conj

( (Pgnettem+j*Qgnetterm)

.

/UO)

;

Edash0=V0+j

"

(Xg.

*

Ig0)

;

Pg0=rea'l

(Edash0.*conj

(Ig0)

)

;

x1

r=angl e(Edash0)

;

% lnit'ial

isation of

state

vector

Pg r=Pg0;

%

Pg_ro1usl.=PgO;

xZ

r=[0

0],;

xtdot

r=[0 0]

,;

x2dot r=[o o]

,;

xldotrpl

rtl=

[0

0J

'

i

x2dotrpl rtl=

J0

0J

'

i

SStep

3

% Here

in

this

example

we have

not

really calculated

YBus

But-sne

% can

write seParate

Program.

yBusdf=

[

1f986-j*35-6301

0

-0.0681+i*5.1661

-j*11.236

0

-0.068l+j*5.

1661

YBuspf=

[

I .3932-j*13.8731

-0.2?I++j*7

.6289

-0.090l+i

*6.0975

%Step

4

0

-0.2214+j*7

.6289

0.5+j

*7

.7898

0

0.

1362-j

*6

.2737f

i

-0.090l+j

*6.0975

0

0.1591=j*6.11681

;

o

f

d]

.o

E

c

.g

o

P

E

o

()

o

c

o

.c)

o

.E

o

(u

EL

cL

go

eo,

OC

o'o/-

L

o

ts

F

=

_o

(U

o

c

.9

o

c

(g

L

F

u?

(5

6

IL

.s

o

% Set

the

va]ues

for

initial

time t(occurance

of

fault)

and

% tc

is time

at

which

fault

is cleared

t=0;

tc=0.08;

tfi nal

=

1 .0;

r=1;

Edash_r=Edash0

;

Edash_rpl

us

1=Edash0

;

while

t

<

tfinal,

%Step 5

Compute

Generator

Powers

using

appropriate

YBus

%

the

YBus chosen

in the

following

step

is set

according

to the current

%

time

if t

<=

tc

YBus=YBusdf;

else

YBus=YBuspf;end

% Note

that

here

we

obtain

the

currents

iniected at

generator

bus

% by multip'lying

the

correspond'ing

row

of

the Ybus

wjth the

ve\tor

of

%

voltages

behind

the

transient

reactances.

This should

also

jnclude

%

slack

bus

voltage

% and

hence

the

entrY

1 aPPears

%

generator

bus

vo'ltages

I=YBus(2:m+1,:)*[1

Edash

rl;

Pg_r=rea1

(Edash_r.*coni

(I)

)

;

jn

the bus

voltage

vector in addition

to

%Step 6

compute

xldot-r

and

xZdot-r

x ldot_r=x2_r;

for k=1:m,

xZdot-r(

k, 1

)

=

(pi

*f/H

(k)

)

*

(PgO

(

k)

-Pg-r

(

k)

)

;

end

%Step

7

Compute

first state

estimates

for t=t(r+1)

x 1_rp1

us

I

=xl_r+x

ldot-r*tsteP

;

xZ_rpl

us

1=x2_r+x2dot_r*tsteP

;

%Step 8 Compute

first estimates

of

E'-r+1

Edash

rpl us1=abs(Edash0)

.*(cos

(x1-rp1

usl)+i*sin(xl-rpl

usl)

)

;

%Step 9 Compute

Pg

for t=t(r+1)

I=YBus(2:m+1,:)*[1

Edash_rp'l

us 1l

;

Pg_rp'l us1=real

(Edash_rp1

us1.*coni

(I)

)

;

%Step

10 Compute

State-denivatives'at

t=t(r+1)

(I'

L

o)

o

c

(\'

=

o)

E

c

at,

IU

-1,

W Mod"rn

po*",.

sy.t.r

An"ryri,

xldot_rplrtl=J0

0J,i

xZdot_rpl

rrl=10

0J

,

i

for

k=1:m,

Iffi

xldot_rpl

usl

(k,

1)

=x2_rp'l

usl

(

k,

l)

;

x2dot.

rpl

usl

(k;

l)=pi

*f/H

(k)

*

(pqg (

%step

11

compute

average

varues

of

state

derivatives

x

I

dotav_r=

(x

ldot_r+x1dot_rp'l

usl)

/

?.0

;

x2dotav_r= (x2dot_r+xZdot_rp1

usl)

/2.0

;

%step

L2

Compute

fjnal

State

estt'mates

fbr

t=t(r+1)

x

l_rp

l us

1=xl_r+xldotav_r*tstep

;

x2_rpl

uslixZ

r+x2dotav_r*tstep

;

%step

13

Compute

final

estimate

for

Edash

at

t=t(r+1)

Edash_rp1

us

1=abs

(Edash0)

.

*

(cos

(xl_rpl

us

1)

+j

*s

i

n

(xt_rp1

usl)

)

;

%Step

14

Print

State

Vector

x?_r=x?_rpl

usl;

xl_r=xl_rp1

usl;

Edash_r=Edash_rp'l

us

I

;

%Step

15

time(r)=1;

for

k=l:m,

ang

(r,

k)

=

(xt_r(k)

*180)

/pi

;

end

t=t+tstep;

r-r+1;

end

plot(time,ang)

Ex.

G.

17

(Example

IZ,II)

%Example

72.11

iE

solved

using

SIMULINK

%

The

code

given

below

should

be run

prior

to

simu'lation

shown

in

Fig.G.6.

cl

ear

al I

globa1

n

ryyr

globalPmfHE

yngg

global

rtd

dtr

%conversjon

factor

rad/degree

91

obal

Ybf

Ydf

yaf

f=50;

ngg=2;

r=5;

nbus=r;

rtd=180/pi;

dtr=pi

/180i

9o

Gen.

gendata=[

1

2

3

Ra

Xd'

H

0

inf

0

0.067

12.00

0

0.100

9.00

l;

EEil$nE

'Fl

88

tt

E

s

3

-8

.=

oll

ct

o

c

.9

U'

c,

6

F

o

.=

.c,

ct

G'

E

=

q

(J

dt

IL

o

a

o

.tt

o

N

o

Il

f

U'

E

o

>

6

II

J

6

iltl

Modern Power

svstem Anatvsts

ydf'I

5.

7986-J

*35.6301

0

-0.0681+i

*5.

1661

-

Appendix

G

E

I

Complex

Product

21

E2

0

-j*11.236

0

-0.0681+J*5.1661

0

0.1362-j*6.27371

i

Yaf=

[

1.3932-j*13.8731

-0.22L4+j*7

.6289

-0.0901+j

*6.0975

-0

.22I4+j*7

.6?89

-0.0901+j

*

6 .097 5

0.5+j*7.7898

0

0.1591-j*6.11681

;

fct= input('fault

clearing

time fct

=

');

&Dampi ng factors

damp2=O.

0;

damp3=0 . 0;

%I

ni

ti

al

generator

Ang'l

es

d2-0,3377*rtdl

d3=0.31845*rtd;

%Initial Powers

Pn2=3.25;

Pm3=2

. 10;

%Generator Internal

Voltages

El.= 1 .0;

E2=1 .03;

E3=1.02;

%Machine Inertia

Constants;

H2=gendat a(2,4)

t

113=gendata

(3,4)

;

&Machl ne Xd'

;

1661=gendata

(2,3)

;

1661=gendata

(3,3)

;

Note : For the simulation

for multimachine

stability

the'two

summation boxes

sum

2 & sum 3

give

the net

acceleating

powers

Po2 and Por. T}ire

gains

of

the

gain

blocks

Gl, G2 G3

and

G4 are

set equal to

pi*fAlZ,

dampZ,

pi*f/Fl3

and

damp3, respectively. The

accelerating

power

P" is

then integrated twice for

each mahcine to

give

the

rotor angles

4

^d

4.

Th" initial conditions for

integrator

biocks integrater

i, integrator

2,integrator

3 anci integrator 4 are set

to

0, d2/r+d, 0 and d3lrtd,

respectively.

The

gain

blocks G5 and G6 convert the

angles

$

and d, into degrees

and

hence

their

gains

are set to rtd.

The

electrical

power

P"2ts calculated

by using

two subsystems

1 and 2.The

detailed diagram

for subsystem

1 is shown in Fig.

G.7. Subsystem

I

gives

two outputs

(i)

complex voltage

Etl6

and

(ii)

current

of

generator

12

which is equal to

Flg.

G.7

Multimachine

transient

Stability

:

Subsystem

1

Yaf

(2,1)

ffi

vodern

power

syster

Rnalysis

E/-61

*Yar

(2,1)

+ Erlq*Yq{z,z) +

ErlErYar

(2,3).

The

switches are

used

to

switch between

prefault

and

post

fault conditions for each machine and their

threshold

values are adjusted to fct, i.e.

the

fault

clearing

time.

cl

ear

% This

%

which

%

lIgll

%

lIszl

%l.l

vol.l

Yo

llgnl

%

lrLll

yo

lrL2l

%l.l

vol.l

%

lILml

I

=

lynl

I

Yn+12

I

Yn+m1

Yfull=

_j*5

0

7*5

0

-j*7.5

j*2.5

j*5 j*?.5

-j*12.51

[row

columns]=sjze

(Yfull)

n=2

YAA=Yfull

(1:n,1:n)

YAB;Yfull

(1:n,

n+1;columns)

YBA=Yful

1

(n+1:

rows, n+L:

col umns)

YBB=Yful

1

(n+1:

rows, n+1

: co1 ums)

%

This

gives

the reduced matrix

Yreduced=YAA-YAB*i

nv

(YBB)

*YBA

matri

x for stabi

'l

i ty

and retains only the

Yln+l

YLn+Z.

YZn+l YZn+Z.

Ynn+1Ynn+2

Program

fi nds

el

imnates

the

I

Yll YL?

iIY21.

Y?r

reduced

I

buses

. Yln

.

YZn

.

Ynn

the

I

oad

studi

es

generator

Y1n+ml

I

V1

Yln+ml

I

V2

tl

buses

2.1

Lirt

=

2.2

0.658

ohmlkm

2.3

L=

F

loR

w^

2r

r

2.4

260.3

Vlkm

2.5

He

=

- .,:I

,

AT/mz (directed

upwards)

n

3nd

2.6

X=

(Xt-Xn)(X2-Xn)

\

*

x2

-zxn

2.7

0.00067

m}J/krlt,

0.0314

V/r<Tl

2.8

0.W44

/-t4C,

ml{lkm,

0.553

/.l40

Vlkm

2.9

0.346

ohmtkm

2.rc

1.48

m

2.ll

0.191

ohmlkm/phase

2.12

0.455

mlllkm/phase

2.13

2.38

m

2.14

(i)

0.557

dtt2

Att4 (ii)

0.633

d2t3

A16

er)

0.746

d3t4

Ar/8

Chapter

3

3.1

n^

_

2

rk

lVl(ln (r/

2D)130"_!n

(DlLr)l_30o)

-,

ru

2h(D/r)ln(r/2nffi

t'tm,

Io

=

2rfqo/90"

A

0.0204

p,Flkrrr

O OlO? ttFl1r'm rn na'l.-^r

/r^,rutr

Lv

uvuLl4l

5,53

x

10{

mhoslkm

8.72

x

l0-3

1fi/xm

ro-7

x

*Vlr,:

-,lt-+n,e|

-,,,)+arlr?]

rs

to

Problems

3.3

0.0096

pFlkm

3.5

3.08

x

iO-s

coulomb/km

3.7

8.54

x

103

ohmslkrn

3.9

71.24

kv

CHAPTER

2

YnZ

Ynn+m

Yn+Lm

Yn+mn+m

Vn

+1

Vn

+1

Vn*m

1

-x

2

3.2

3"4

3.6

3.8

Chapter

4

4.1

12

kV

W

Modrrn

Po*"r.

syrt"t An"lv.i.

Chapter

5

5.1

(a)

992.75 kW

(b)

No

solution

possible

5.2

At

-0.91I.5o,

B'=239.9166.3",

C'=

0.001 1102.6,

Dt

=0.851I.96

(b)

165.44

ky,0.2441-28.3"

kA,

0.808 lagging,

56.49 MW

(c)

7O.8Vo

(d)

28.L5Vo

5.4

(a)

273.5

MVA

(b)

l,IT4

A

(c)

467.7 MVA

5.5 133.92

kV

,

23.12

MW

5.6 202.2

kV

5.7 At

x

=

0,

irr

=

0.3L4 cos

Q,st

-21.7"),

irz=0.117

cos

(utt

+ 109"),

At

x

=

200

km, it

=

0.327

cos

(c..r/

-

9.3o), irz

=

0.t12 cos

(,*t

+

96.6)

5.8 135.817.8o

kV,

0.138115.6o

kA,

0.99

leading,

55.66 MW,

89.8Vo,

373.L1

-

t.5o,

3,338

km, 1,66,900

km/sec

5.g v

-

12g.3172.60,

Y'

-

0.000511g9.5".

2

5.10 7.12",

pfr

=

0.7 lagging,

pfz=

0.74 \agging

5.11 47.56

MVAR

lagging

5.I2

10.97

kV, 0.98 leading,

-

0.27Vo,86.2Vo

5.13 51.16

kV,

38.87 MVAR

leading,

40

MW

5.I4 238.5

kV,

P,+

jQ, =

53

-

jI0,

pf

-

0.983 leading

5"15 17.39

MVAR

leading,

3.54 MW

Chapter

6

6.1 For this

network

tree is

shown in Fig.

6.3a;A is

given

by Eq.

(6.17).

The

matrix

is not unique.

It depends

upon

the orientation

of

the

elements.

6.2

V)

=

0.9721-

8.15"

6.3 V)

=

1.26 l:74.66"

6.a

@)

IE

(b)

j0.3049

j0.1694

j0.1948

j0.3134

0.807

_

js.6s

0.645

_

j4,517

0.968

-

j6.776

0.968

_

j6.776

0.880

_

j6.160

6.5

Pn

=

-

0.598

pu,

PB

=

0.2

pu,

pzs

=

0.796

pu

Qn=

Qzr

=

0.036

pu,

Qn=

0.004

pu,

ezz=

0.064

pu

6.6

(a)

Pn

=

-

0.58 pu,

P*

=

0.214

pu,

pzl

=

0.792

pu

Qn

=

-

0.165

pu,

ezt

_

0.243

pu,

en

=

0.204

pu

Qy

=

-

0.188

pv,

Qzt

=

0.479

pu,

en=

-

0.321

pu

(b)

Pn

=

-

0.333

Pu,

Pzz=

0.664

pu,

Pr:

=

-

0.333

pu

'

on=

Qzr

=

0'011pu,

QB=

Qzr

=

0.011

pu,

ezt=

en=g.g44

pu

(c)

tz}4

I l-

-ito

s/s3"

i5

I

tul

lst87"

-ilo

js

I

J

L

js

js _jl0j

f-iro.rors

j5.o5o5

js

6.7

(a)

(i)

|

;s.osos

-j10

js

L

j5

js

_j10

t

2

3

4

--:-

)

6

7

I

9

U

I

0

I

0

001

-l

00

l-1

0

0-l

0

-l

0l

U

0

I

0

U

Qr

0

I

fh\/i\ D

-n<nn-..

n n A^A

\.r,,

\i,,

i

12:

v.iJVU

pu,

ts13

=

U.LUZ

pu,

pzZ

=

0.794

pu

Qn=

0.087 pu,

Qzt

=

-

0.014

pu

Qn=

QzF

0.004

pu,

Qzz

=

Qsz

=

0.064

pu

//::\ I) t\ .oE

\u,,

f

12:-

u.ooJpu,

Pn

=

0.287 pu,

Pzt

=

0.711

pu

Qn

=

0.047

Pu,Qn

=

0.008

ptr,

Qzt

=

0.051

pu

6.8

Vrt

=

I.025

-

y0.095

=

l.AZ9/.-

5.3. pu

-l

1

0

1

0

4

Mooern

power

slrstem

nnatysii

Chapter

7

7.1

Rs 22.5/hr

7.2

(a)

Pcr

=

140.9

MW,

159.1

MW

7.3

7.4

7.5

7.6

7.8

7.7

S4VlIl$

=

(i)

Gen

A will

share

more

load

than

Gen

B

(ii)

Gen

A and

Gen B

will

share

load

of

po

each

(iii)

Gen

B

will

share

more

load

than

Gen

A.

Pcr

=

148

MW,

Pez,

=

L42.9

MW,

Pct

=

109.1

MW

(dcldPd=0.175Pc+23

(a)

Pq

=

138.89

MW, Pcz

=

150

MW,

Po

=

269.6

MW

0)

Pcr

=

310.8

MW,

Pcz

=

55.4

MW

(c)

For

part (a):

Cr

=

Rs

6,465.141hr

Forpart

(b):

Cr

=

Rs

7,708.15/hr

Bn

=

0.03387 pu

or

0.03387 x

ITa

MW-l

Brz

=

9.6073 x

10-5

pu

or

9.6073 x

10-7

MW-t

Bzz

=

0.02370

pu

or

0.02370

x

10-2

MW-r.

Economically

optimum

UC

Load

(MW)

Unit Number

1234

0-4

4-8

8-r2

t2-16

r6-20

20-24

Optimal

and

secure

UC

Period

Unit

Number

t234

04

4-8

8-12

I2_16

l6-20

20-24

7.9

Total

operating

cost

(both

units

in

service

for24

hrs)

=

Rs

1,47,960

Total

operating

cost

(unit

tr

put

off

in light

load

period)

=

Rs

1,45,g40

1111

r110

1100

1110

1000

1100

20

l4

6

l4

4

10

1111

1'1

1

0

1

0

0.

11,1

0

1100

rt00

RL.,-iiiis;E:

lit+i(Atlldt

I--

8.1

Load

on

123

MW,

Load

on

G2

=

277

MW,

50.77

H)..

2

8.2

Af

(t)

=

-

0.029

-

0.04ea.58'

cos (1.254t

+

137.g")

8.3

1/(50K)

sec

9.4

aPtie.

|

-

_

ll

R2)

an

(K

iz+D0

/ K

p,

1

*

K

ipl

+l

/ Rt)+ (K,

r

8.5

APti.,r(s)

=

-

tOOf

O'ztt

t0.9t

L

80s5

1

System

is

found

to

be

unstable.

Chapter

9

9.r

9.2

9.3

it

=

3.14

sin

(314t

-

66.)

+

2.97e

sor,

;"*

=

5A

(a)

81"

(b)

-

9'

(1)

Ie

=

2.386

kA,

.IB

-

9.4

8.87

kA,

4.93

kA

9.6

6.97

kA

9.7

(a)

0.9277

kA

(b)

1.312

kA

(c)

(e)

0.1959

kA

8.319

kA

1.75

kA

(ii)

IA

=

4.373

kA,

IB

=

|.ZS

ta

9.5

26.96

kA

1.4843

kA

(d)

1.0205

kv,

53.03

MVA

9.9

2.39

pu

9.8

9.10

9,12

132.1,

47.9;

136.9,

45.6

gJt

0.6 pu

{

-

j8,006

pu,

1{:

=

-

74.004

pu

Chapter

10

10.1 (i)

1.7321270"

(ii)

zl0"

(iii)

1

.7321150.

(iv)

tl2r0"

lO.2

IA=

jl.l6

pu,

Vtn

=

1.171109.5"

pu

ll

-

n O<2 ./ 4.< Ao --- Y'

v

BC

-

v.zirz_-

oJ.+-

pu,

VCB

=

U.9952_Il3.lo

pu.

19

3

Vor

=

l97.Vl-

3.3"

V,

Voz

J

20.Zll51.lo

V,

V"o

=

21.61110.63"

V

10.4

lor

=

19.23_1.

3V

A,

Ioz

=

Ig.23lISV

A,1oo

=

0

A

10.5

1,,=27-87,/3ff

A t,--'t?./- AAo.lo A r n a

^r

^AZ

-

LJ1-

a-'.ZJ

Il,

lA0

=

U

A

Iobr

=

16.1

A,

Iab2

=

7.5/_-75"

A,

Iob1=

l.,S/lS

e,

10.6

Io

=

16.16

+

j1.335

A,

Iu=

_

9.24

_"1\O.AO

e,

Ic

-

6.93

+

j9.32

A,

lVN,l

=

lVosl

=

40.75

y

t0.7

1,500.2

w

I

l

ffi4

Modern

Power

System

Anal'sis

Chapter

11

1 1.1

-

i6.56

kA,

lV6,l

-

12.83

kV,

lvobl

=

6.61

kV,

ly"rl

=

6'61

kV

=

I

=

-2./1

(b)

V*=

Vo,

=

0'816

pu, 116l

=

llrl

=

5'69

pu

11.3

(i)

-

i6.2.5

(ii)

-

4.33

(iii)

6'01

(iv)

-

/5

pu

In

order

of

decreasing

magnitude

of

line

currents

the

faults

can

be

listed

AS

(a)

LG

(b)

LLG

(c)

3-Phase

(d)

LL

Il.4

0.1936

ohm,

0.581

ohm,

-

4.33

pu,

j5

pu

11.5(a)3.51pu(b)Vt,=I.|91_159.5"po,V,=I,681|29.8"pt1

(c)

0.726

Pu

11.6

Iu=-Ir=-2.887Pu

||.7

(a)

Iv

=

_

5.79

+

j5.01

kA,

18

=

5.79

+

j5.01

kA,

,IG

= j10.02

kA

(b)

/n

-

IY

=

-

6'111

kA,

Ic=

0

1 1.8

In,

=

0

Io^

=

-

73'51

Pu

Ius

=

-

72.08

pu

Ib^=

-

jl.Z

Piu

IrB

=

72.08

pu

I,^

=

-

jl.Z

Pu

11.9

5,266

A

11.10

j2.0

pt

11.1

1

/

=

-

j6.732

Pt,

Io(A)

=

-

i4.779

pu,

/a(A)

=

-

i0.255

po, 1"

(A)

-

j0'255

prt

l|lz

0.42

ptt,

-

j9'256

Pu

11.13

-ill.l52

pu,

-i2.478

pu,

-Jl

-239

pn

'1.14

4.737

Pu,

1

Pu

r 1.15

f,

=

-

i12.547

pu, Ifrr(b)

=

-

i0.0962

pu

Chapter

12

l2.I

4.19

MJA4VA,

0.0547

MJ-sec/elec

deg

12.2

4.315

MJA4VA

12.3

40'4

MJiTVIVA

12.4

140.1

MW,

130.63

MW,

175.67

MW

r2.5

72.54

MW

!2.7

L27.3

MW

12.6

4

=

58'

12.8

53'.

We

need

to

know

the

inertia

constant

M

to

determine

/.'

ABCD

constants

61,7

for various

simple

networks

618

in

power

flow

equations

lS9

measurement

of

621

of

networks

in

series

and

parallel

620

AC

calculating

board

lB4

Accelerationfactor

207

Accelerating

power

462

Acid

rain

16

Adjoint

matrix

609

Admittance

matrix

(see

Bus admittance

matrix)

Advanced

gas

reactor

(AGR)

19

Algorithm

for

building

the

Zru,

355

for

load

flow

solution

by

GS

method

205

by NR

method

214

by FDLF

223

for

optimum generation

scheduling

262

for

optimal

hydrothermal

scheduling

280

for optimal

loading

of

generators

on

a

bus 246

for

optimal

load

flow

solution

273

for

short

circuit

current

computation

343

for

short

circuit

studies

349

for static

state

estimation

of

power

systems

540

for

transient

stability,

analysis

of

large

system

4gS

Alternator (sea

Synchronous

machines)

Aluminum

conductor

steel

reinforced

(ACSR)

conductors

52

A *aa a^-+-^1 ^-^- / a ^F\ 4A f,

Arvc

Lrrllll(I

Ellul

\f\\-.8,

JU+

Armature

leakage

reactance

331

Armature

reaction

109,

110,

330

Attenuationconstant

143

Augmented

cost

function

279

Automatic generation

conirol

(see

Load

frequency

control)

Automatic

voltagc

controi

318

Index

12.9

The

system

is

stable

l2.Il

The

system

is

unstable

12.13

The

system

is

stable

12.10

70.54",

0.1725

sec

t2.t2

63.36"

12.14

The

system

is

stable

B-coefficients

261.267

Bad

data

detection

547

Bad

data

detection

547

identification

547

suppression

548

treatment

546

Base

load

3

Base

value

99

Bharat

Heavy

Electricals

Lrd

(BHEL)

14,31

Boiling

water

reactor

19

Breaking

resistors

for

improving

stabiliry

499

Branch

190

Breakers (see

Circuit

breakers)

Brown,

H.

E.

368

Bundled

conductors,

\

capacitivereactance

92

inductive

reactance

68

Bus

generator

198

Ioad

198

PQ

198

PV

198

reference

198

slack

Bus

admittance

matrix

188

formulation

of

187.

189

Bus

impedance

matrix

building

algorithm

355

for

unsymmetrical

fault

analysis

416

in crra*^+-l^^l f^..fr ---!---' -Er

rrr

DJrrrrtlsttrucu

latuil.

al|alysls

JJ

I

Bus

incidence

matrix

I92

Capacitance

^^f^--1^-i--- t -t

u4ruuliluon

oy

memoo

oI

modrlred

geometric

mean

distances

9l

effect

of

earth

on

83

effect

of

non-uniform

charge

distribution

79

effect

of

stranded

conductors

79

line-to-line

78

line-to-neutral

79

12.15

The

system

is

unstable

for

both

three

pole and

single

pole switching

tndex

of

parallel

circuit

three-phase

lines

88

of

three-phase

lines:

with

equilateral

spacing

80

with

unsymmetrical

spacing

81

a two-wire

line

(see

also

Reactance)

78

Central

Electricity

Authority (CEA)

Zg

Charpcteristic (surge)

impedance

l4l

qf

.lines

and

cables

145

Chaiging

current

76, 180

Circle

diagrams

167

Circuit

breakers

327, 329

autoreclose

460

rated

intemrpting

capacity

of

344

rated

momentary

current

of

344

selection

of

344

Circulating

currents

389

Cogeneration

15

Coherent group

291,303

Compact

storage

scheme

189,628

Comparison

of angle

and voltage

stability

592

Compensation

series

779,498,

,Sg

shunt

179,498,562

Compensator

combined

TCR

and TSC

.564

shunt

562

static

synchronous

series

(SSCC)

571

Complex power

105

Composite

conductors

capacitance

9l

inductance

54

Computational

flow

chart

for

load

flow

solution

using

FDLF

227

GS method

20S

NR

method

224

Conductors

ACSR

52

Bundled

52,68

expanded

ACSR

52

iypes

5 i

Conductance

45

Constraints

equality

272,

632

inequality

632

on

control

variables

274

on

dependent

variables

274

for

hydro-thermal

scheduling

.

problems

277

Contingency

analysis

51

rankins

520

screening

520

selection

515

'

direct

method

515

indirect

method

515

Contingency

analysis

by dc

model

520

by

distribution

factors

521

modelling

for

5I4

Contingency

evaluation

(see

contingency

analysis)

Control

area

291,303,306

by

transformers

23I

integral

304

isochronous

305

of

voltage

profile

230

of

WAfiS

and

VARS

along

a

transmission

line

230

optimal

310

parameters

199,272

proportioiral

pft"s-fuBral

303

suboptimal

318

Control

area

concept

303

Control

strategy

303

Control

variables

199,

632

Control

vector

199

Controller

interline

power

flow

(IpFC)

571

interphase

power (IPC)

572

unified power

flow

(UPFC)

571

Converters

567

Ccxrrdinatisn

equations

2fr

Corona

52,68

Cost function

251,270

i,t^e-^^ 1nn

\-ull(i('

L>V

Covariance

matrix

535

Critical

clearing

angle

and

time

467

Current

distribution

factors

265

Current

limiting

reactors

346

DAC

(Distribution

Automation

and

Control

System)

634

Damper

winding

438

DC

network

analyser

39

Data

acquisition

systems

(DAS)

636

Digiral

LF

controllers

322

Direct

axis

reactance

llg

Disturbance

parameters

271,

272

Distribution

factors

generation_shift

SZ0

Iine-outage

520

Diversity

factor

4

Dommel,

H.

W.

270

Doubling

effect

3Zg

Double

line-to-ground

(LLG)

fault

404

Driving

point

(self)

admittance

lg7

Dual

variables

279,632

Dynamic

programming

applied

to

unit

commitment

251

Economic

dispatch

(see

also

optimum

generation

scheduling)

306

Electricity

Board

Zgl

Elements

189

Energy

conservation

3l

Energy

control

centre

637

!MS.(Energymanagement

system)

634

Energy

sources

conventional

13

hydroelectric

power

generation

l7

nuclear

power

stations

lg

thermal

power

stations

13

biofuels

28

biomass

28

gas

turbines

16

geotbermal

power

plants

24

magnetohydrodynam

i

c

(MHDI

generation

23

OTEC

27

renewable

Zs

solar

energy

26

wave

energy

27

wind

power

stations

25

Energy

storage

Zg

fuel

cells

29

hydrogen

Zg

pumped

storage

planf

lg

secondary

batteries

2g

Equal

area

criterion

461

Equal

incremental

fuel

cost

criterion

246

the

medium

length

line

nominal_

zr

representation

l3g

nominal-Trepresentation

137

short

transmission

line

l}g

synchronous

generator

llg

Error

amplifier

3lg

Error

signat

319

Estimation

methods

least

squares

532

weighted

least-squares

533

Estimation

of periodic

components

5g

l

Exact

coordination

equation

26I

Expectation

532

External

system

equivalencing

545

Facrors

affecting

srability

496

FACTS

controllers

569

Failure

rate

255

Fast

breeder

reactor

22

Fast

decoupled

load

flow

223

Fasr

valving

49,9

Faults

balanced

(see

symmetrical

fault

analysis)

calculation

using

Zsus

351,417

open

conductor

414

unbalanced

series

type

397

shunt

type

397

Ferranri

effect

150

Field

erciretion

l0g

Field

rriading

32g

Flat

voltage

starr

205,209,21g

Flexible

AC

transmission

systems

(F-ACTS)

566

Fluidized-bed

boiler

15

Flux

linkages

external

49

internal

46

of

an

isolated

current

carrying

conductor

46

,Fly

ball

speed

governor

Zgz

rorecastlngmethodology

577

Fortescue,C.L.

370,3t6

Frequency

bias

309

Fuel

cost

ofgenerators

243

Kothari D.P., Nagrath I.J. Modern Power Systems Analysis

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